Answer :
Certainly! Let's determine if [tex]\(x = \frac{1}{2}\)[/tex] is a zero of the polynomial [tex]\(p(x) = 2x + 1\)[/tex].
### Step-by-Step Solution:
1. Identify the polynomial:
The given polynomial is:
[tex]\[ p(x) = 2x + 1 \][/tex]
2. Substitute the given value [tex]\(x = \frac{1}{2}\)[/tex] into the polynomial:
To find out if [tex]\(x = \frac{1}{2}\)[/tex] is a zero of the polynomial, we need to substitute [tex]\(x = \frac{1}{2}\)[/tex] into the equation and see if the result is zero.
Let's do the substitution:
[tex]\[ p\left(\frac{1}{2}\right) = 2 \left(\frac{1}{2}\right) + 1 \][/tex]
3. Simplify the expression:
Simplify the right-hand side of the equation step-by-step:
[tex]\[ 2 \left(\frac{1}{2}\right) = 1 \][/tex]
Now, add the remaining term:
[tex]\[ 1 + 1 = 2 \][/tex]
4. Evaluate the polynomial:
After substituting [tex]\(x = \frac{1}{2}\)[/tex], we get:
[tex]\[ p\left(\frac{1}{2}\right) = 2 \][/tex]
5. Determine if the result is zero:
A zero of the polynomial [tex]\(p(x) = 2x + 1\)[/tex] would ensure that:
[tex]\[ p(x) = 0 \][/tex]
However, we found that:
[tex]\[ p\left(\frac{1}{2}\right) = 2 \][/tex]
### Conclusion:
Since [tex]\(p\left(\frac{1}{2}\right) = 2\)[/tex] and not zero, we can conclude that [tex]\(x = \frac{1}{2}\)[/tex] is not a zero of the polynomial [tex]\(p(x) = 2x + 1\)[/tex].
So, the answer is:
[tex]\[ p\left(\frac{1}{2}\right) = 2 \quad \text{and} \quad \frac{1}{2} \text{ is not a zero of the polynomial.} \][/tex]
### Step-by-Step Solution:
1. Identify the polynomial:
The given polynomial is:
[tex]\[ p(x) = 2x + 1 \][/tex]
2. Substitute the given value [tex]\(x = \frac{1}{2}\)[/tex] into the polynomial:
To find out if [tex]\(x = \frac{1}{2}\)[/tex] is a zero of the polynomial, we need to substitute [tex]\(x = \frac{1}{2}\)[/tex] into the equation and see if the result is zero.
Let's do the substitution:
[tex]\[ p\left(\frac{1}{2}\right) = 2 \left(\frac{1}{2}\right) + 1 \][/tex]
3. Simplify the expression:
Simplify the right-hand side of the equation step-by-step:
[tex]\[ 2 \left(\frac{1}{2}\right) = 1 \][/tex]
Now, add the remaining term:
[tex]\[ 1 + 1 = 2 \][/tex]
4. Evaluate the polynomial:
After substituting [tex]\(x = \frac{1}{2}\)[/tex], we get:
[tex]\[ p\left(\frac{1}{2}\right) = 2 \][/tex]
5. Determine if the result is zero:
A zero of the polynomial [tex]\(p(x) = 2x + 1\)[/tex] would ensure that:
[tex]\[ p(x) = 0 \][/tex]
However, we found that:
[tex]\[ p\left(\frac{1}{2}\right) = 2 \][/tex]
### Conclusion:
Since [tex]\(p\left(\frac{1}{2}\right) = 2\)[/tex] and not zero, we can conclude that [tex]\(x = \frac{1}{2}\)[/tex] is not a zero of the polynomial [tex]\(p(x) = 2x + 1\)[/tex].
So, the answer is:
[tex]\[ p\left(\frac{1}{2}\right) = 2 \quad \text{and} \quad \frac{1}{2} \text{ is not a zero of the polynomial.} \][/tex]
